Question: Divide the following complex numbers. $ \dfrac{-7+3i}{5+2i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${5-2i}$ $ \dfrac{-7+3i}{5+2i} = \dfrac{-7+3i}{5+2i} \cdot \dfrac{{5-2i}}{{5-2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-7+3i) \cdot (5-2i)} {(5+2i) \cdot (5-2i)} = \dfrac{(-7+3i) \cdot (5-2i)} {5^2 - (2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-7+3i) \cdot (5-2i)} {(5)^2 - (2i)^2} = $ $ \dfrac{(-7+3i) \cdot (5-2i)} {25 + 4} = $ $ \dfrac{(-7+3i) \cdot (5-2i)} {29} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-7+3i}) \cdot ({5-2i})} {29} = $ $ \dfrac{{-7} \cdot {5} + {3} \cdot {5 i} + {-7} \cdot {-2 i} + {3} \cdot {-2 i^2}} {29} $ Evaluate each product of two numbers. $ \dfrac{-35 + 15i + 14i - 6 i^2} {29} $ Finally, simplify the fraction. $ \dfrac{-35 + 15i + 14i + 6} {29} = \dfrac{-29 + 29i} {29} = -1+i $